0
$\begingroup$

Consider a linear subspace Y of a symplectic vector space (V,w). Its symplectic orthogonal is defined by

$$Y^O=\{v\in V|w(v,u)=0~ \forall ~u \in Y\}$$

Now, apparently, we must have

$$(Y^O)^O=Y$$

for any Y, but this confuses me a bit. Here is what I'm thinking:

Let's say V is for simplicity 6 dimensional and has a basis of unit vectors $e_1,e_2,e_3,f_1,f_2,f_3$ such that $$w(e_i,f_j)=\delta_{i,j} ~~,~~ w(e_i,e_j)=w(f_i,f_j)=0$$

Then Y might be e.g. a subspace with basis $e_1,e_2,f_1$. Following the definition above, only $e_3,f_3$ would have vanishing symplectic product with all three of these vectors, so that $e_3,f_3$ are a basis of $Y^O$. But then by the same logic $(Y^O)^O$ would have a basis $e_1,e_2,f_1,f_2$, which is more than the original 3 vectors and cannot be equivalent to Y. So it seems $Y\neq (Y^O)^O$ after all.

Is there some mistake in the above logic? How should the argument properly go?

$\endgroup$

1 Answer 1

0
$\begingroup$

The problem with your example is that you've missed another independent element of $Y^O$: $e_2$ is also orthogonal to $Y$, so that $e_2, e_3, f_3$ is a basis of $(Y^O)^O$.

I think the statement only holds in the finite-dimensional case, and if we assume that $V$ is finite-dimensional then we can use a dimension-counting argument to prove it. Suppose that $V$ is $k$-dimensional and $Y$ is $n$-dimensional, and choose a basis $b_1, \ldots, b_n$ of Y. Use this to define a map $\phi : V \to F^n$ (where $F$ is the field over which $V$ is defined) by $$ \phi(v) = (w(b_1, v), \ldots, w(b_n, v)) $$ By the non-degeneracy of $w$, this map is surjective. Its kernel is therefore $k-n$ dimensional, and is of course equal to $Y^O$. Now define a similar map $\phi^O$ using a basis for $Y^O$, and note that $Y \subset \ker(\phi^O)$, and $\dim(Y) = \dim(\ker(\phi^O))$. Therefore $Y = \ker(\phi^O) = (Y^O)^O$.

There might be a cleaner argument...

$\endgroup$
6
  • $\begingroup$ I thought $w(e_2,f_2)\neq 0$, so $f_2$ does not fit the definition of vectors that may be in $Y^O$ while $e_2$ is in Y? Cluld you elaborate a bit on why $f_2$ is allowed within $Y^O$? $\endgroup$
    – hms
    Commented Jul 27, 2019 at 4:25
  • $\begingroup$ I think you misread :-). It is $e_2$, not $f_2$, that is in $Y^O$. $\endgroup$
    – Rhys
    Commented Jul 27, 2019 at 4:26
  • $\begingroup$ Oh, you are right! Thank you so much, that makes a lot of sense! $\endgroup$
    – hms
    Commented Jul 27, 2019 at 4:27
  • $\begingroup$ If $w$ was in fact degenerate, wouldn't $\phi$ still be surjective? $k$ would just include a number of decoupled dimensions? $\endgroup$
    – hms
    Commented Jul 27, 2019 at 4:51
  • $\begingroup$ Do we obtain $dim(ker(\phi^O))$ by direct inspection of an explicit standard (up to isomorphism) symplectic basis, or is there maybe a sleek argument how to get that number? $\endgroup$
    – hms
    Commented Jul 27, 2019 at 4:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .